Chemical Kinetics — NEET UG practice

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Sample questions with solutions

Q1 · 2026

Given below is an expression for the rate constant of a first-order reaction occurring at a certain temperature, T(K)\mathrm{T(K)}.

ln⁡k=14.34−1.25×104T\ln k = 14.34 - \frac{1.25\times10^{4}}{T}

The energy of activation in kcal mol−1\text{kcal mol}^{-1} for the reaction is :

(Given: kk in s−1\text{s}^{-1}, R=1.987 cal mol−1K−1R = 1.987\ \text{cal mol}^{-1}\text{K}^{-1})

  • A.

    24.84

  • B.

    14.34

  • C.

    18.63

  • D.

    12.42

Answer: A
  1. Compare the given equation with the Arrhenius equation in its logarithmic form, since matching the two expressions lets us identify EaE_a.
ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}
  1. Matching the coefficient of 1T\frac{1}{T} in both expressions:
EaR=1.25×104\frac{E_a}{R} = 1.25\times10^{4}

Therefore:

Ea=1.25×104×RE_a = 1.25\times10^{4}\times R
  1. Substitute R=1.987 cal mol−1K−1R = 1.987\ \text{cal mol}^{-1}\text{K}^{-1} to get EaE_a in cal/mol, then convert to kcal/mol.
Ea=1.25×104×1.987=24.84×103 cal mol−1=24.84 kcal mol−1E_a = 1.25\times10^{4}\times 1.987 = 24.84\times10^{3}\ \text{cal mol}^{-1} = 24.84\ \text{kcal mol}^{-1}

Hence, the answer is A, 24.84.

Q2 · 2026

2A→kB2A \xrightarrow{k} B is a zero-order reaction, where k=1.0 mol L−1min−1k = 1.0\ \text{mol L}^{-1}\text{min}^{-1}. If the initial concentration of A is 2 M, then the time taken to complete 75%75\% of the reaction will be

  • A.

    2.0 min

  • B.

    1.5 min

  • C.

    0.75 min

  • D.

    1.0 min

Answer: C
  1. Write the rate law for the reaction, keeping the stoichiometric coefficient of A in mind, since the rate is defined per mole of reaction.
−12d[A]dt=k-\frac{1}{2}\frac{d[A]}{dt} = k
  1. Rearrange this differential form into an equation for time in terms of concentration change.
t=[A]0−[A]t2kt = \frac{[A]_0 - [A]_t}{2k}
  1. Given [A]0=2 M[A]_0 = 2\ \text{M}, 75% completion means 75% of A is consumed, so [A]t=0.5 M[A]_t = 0.5\ \text{M}.

Therefore, substituting the values:

t=2−0.52×1=0.75 mint = \frac{2 - 0.5}{2\times 1} = 0.75\ \text{min}

Hence, the answer is C, 0.75 min.

Q3 · 2026

Match List I with List II :

List-I (Order of Reaction)List-II (Unit of Rate Constant)
A.Zero orderI.mol−1L s−1\text{mol}^{-1}\text{L s}^{-1}
B.First orderII.mol−2L2s−1\text{mol}^{-2}\text{L}^{2}\text{s}^{-1}
C.Second orderIII.s−1\text{s}^{-1}
D.Third orderIV.mol L−1s−1\text{mol L}^{-1}\text{s}^{-1}

Choose the correct answer from the options given below :

  • A.

    A-IV, B-II, C-I, D-III

  • B.

    A-IV, B-III, C-I, D-II

  • C.

    A-IV, B-III, C-II, D-I

  • D.

    A-I, B-II, C-III, D-IV

Answer: B
  1. Recall the general formula for the unit of the rate constant of an nthn^{th} order reaction, since this directly gives the unit for each order listed.
Unit of k=(molL)1−ns−1\text{Unit of } k = \left(\frac{\text{mol}}{L}\right)^{1-n} s^{-1}
  1. For zero order (n=0n=0):
k=mol L−1s−1k = \text{mol L}^{-1}\text{s}^{-1}

This matches IV.

  1. For first order (n=1n=1):
k=s−1k = \text{s}^{-1}

This matches III.

  1. For second order (n=2n=2):
k=mol−1L s−1k = \text{mol}^{-1}\text{L s}^{-1}

This matches I.

  1. For third order (n=3n=3):
k=mol−2L2s−1k = \text{mol}^{-2}\text{L}^{2}\text{s}^{-1}

This matches II.

So, the correct matching is A-IV, B-III, C-I, D-II. Hence, the answer is B.

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